Posted by Anonymous on Thursday, November 27, 2008 at 12:20pm.
You want to remove M C (delta T) = 300 g*1.00 Cal/(g C) * 29 C = 8700 calories from the coffee. If m is the mass of ice you add, the heat absorbed by the ice while melting and incresing in temperature from -20 to +58 C is
m*Cice*(20) + mCwater*58 + m*80 Cal/g
The specific heat of ice, Cice = 0.51 Cal/g C
8700 cal = 10.2 m + 58 m + 80 m = 148.2
Solve for m
q1 = heat to move ice from -20 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). [Tf will be 0 and Ti will be -20 but watch that sign since it will change.
q2 = heat to melt the ice.
q2 = mass ice x heat fusion.
q3 = heat to move ice from zero to 58.
q3 = mass ice x specific heat water (its now a liquid) x (58-0).
q4 = loss of heat to move 300 mL water from 87 to 58.
q4 = mass water x specific heat water x (58-87).
q1 + q2 + q3 + q4 = 0
You have only one unknown in all of the above which is mass ice. Solve for that. Check my thinking. I won't be on for the next 6 hours or so.
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